Predicting the response of olfactory sensory neurons

Predicting the response of olfactory sensory neurons

to odor mixtures from single odor response

A. Marasco, A. De Paris, A. Migliore

Appendix A.1

In [1]-[2] and [3] two di¤erent models were proposed, to take into account the responses F _U and F _{U +V} to concentrations of an odorant U and of a binary mixture U + V , respectively. In this appendix we discuss some features of those models, in view of the empirical properties mentioned in the main text.

To turn those experimental qualitative descriptions into rigorous mathematical de…nitions, one should take into account that they involve terms that are not necessarily consensual in the literature, or that cannot be easily expressed into mathematical relations. For this reason we will focus the analysis on the three basic behaviours of suppression, inhibition and synergy (see [5],[2]).

In most cases, the responses F U , F V to single odorants, and F U +V to the mixture, are compared at the same concentration. This means that if a mix- ture is performed at a certain ratio r > 0, then is U = rV , and the response F U +V (U; V ) = F U +V (rV; V ) is relative to the concentration C := U + V = (1 + r)V . Henceforth, it is compared with the responses F U (C) and F V (C). In this context, we say that inhibition happens when

F _{U +V} (U; V ) < minfF U (C); F _V (C)g ; synergy when

F _{U +V} (U; V ) > maxfF U (C); F _V (C)g;

and suppression when F _{U +V} (U; V ) is (strictly) intermediate between F _U (C), F _V (C). Hence we ignore the comparison of F _{U +V} (U; V ) with the sum F _U (U ) + F V (V ).

The above interpretation will mainly be used in Appendix A3. In the main text we explain that a slightly di¤erent interpretation is also worthy of consid- eration. It basically consists in considering V = C=(1 + r) in place of C, in the response to a mixture. These interpretations do not con‡ict in the asymptotic region.

Rospars et al. models

In [1] and [2], the authors proposed the following models

F U = F _{M U}

1 + ^K _U

ⁿ ; F U +V (U; V ) = F M U

U K U

n

+ F M V

V K V

n

1 + U

K U n

+ V

K V

n (1)

where n is a Hill coe¢ cient, K _U ; K _V ; F _{M U} and F _{M V} are constants (see [2]).

Cruz and Lowe models

In [3] the authors, instead of (1), proposed the following models

F _U = F _max

1 + 1

n

U

1 + K U

U

n ; F _{U +V} (U; V ) = F _max 1 +

2 6 6 4

1 + U K U

+ V K V

U

U K U

+

V K V

3 7 7 5

n ;

(2) where n is a Hill coe¢ cient, F max is the maximal response,

and V are the e¢ ciencies of activation of transduction by each odorant.

Equations (1) 2 and (2) 2 are both functions of two compounds in concen- tration U and V: However, following the experimental protocol, it is useful to analyze them under the condition

U = rV; (3)

where r is a positive constant.

Rospars et al. binary mixture model In [2] is proved that Eq. (1) ₂ becomes

F _{U +V} (rV; V ) = F _{M p} 1 + _(1+r)V ^K

n ; F _{U +V} (M ) = F _{M p} 1 + ^K _M

n ; (4)

where M = U + V and

F _{M p} = F _{M U} r ⁿ K _V ⁿ + F _{M V} K _U ⁿ

(K _U ⁿ + r ⁿ K _V ⁿ ) ; K _p = (1 + r) K _U K _V

s 1

r ⁿ K _V ⁿ + K _U ⁿ : (5) Then, F M p is the asymptotic value of function F U +V (rV; V ) ; the concentration of V (resp. of M ) at half maximum response F M p =2 is K p = (1 + r) (resp. K p ).

Moreover, the asymptotic value of function F _{U +V} is always located between the single-odorant curves. In fact, from (5) for all r we obtain

min (F M U ; F M V ) F M p max (F M U ; F M V ) : (6)

Cruz and Lowe binary mixture model

Analogously, if the concentrations U and V of the odorants are such that U = rV Eq. (2) ₂ becomes

F _{U +V} (U; V ) = F _{U +V} (rV; V ) = F _max 1 + 1

n 1 + K V

n (7)

where the parameters = (r;

; V ; K U ; K V ) ; K = K (r; K U ; K V ) assume the form

= ^U rK _V + _V K _U K U + rK V

; K = K _U K _V K U + rK V

: (8)

In this case, the asymptotic value of function F U +V (rV; V ) is F max

1 + 1

n

=

n F max

1 + ⁿ ; (9)

and the concentration of V at half maximun response

n F max

2 (1 + ⁿ ) is K

p

2 + ⁿ 1 (10)

From (8) we obtain

min ( _U ; _V ) max ( _U ; _V ) and owing to (9) for all r we have

min 0 B B

@ F max

1 + 1

n U

; F max

1 + 1

n V

1 C C A

F max

1 + 1

n

max 0 B B

@ F max

1 + 1

n U

; F max

1 + 1

n V

1 C C

A : (11)

Then, owing to (6) and (11), both the models (4) and (7) cannot reproduce

mixtures exhibiting synergy or inhibition in their maximal response.

Appendix A.2

Model for binary mixture

In order to …nd a model able to describe each type of interaction among two single odorant it is necessary to expand the space of allowed Hill functions. To this purpose, we assume that the Hill coe¢ cient in the mixture model (2) 2 can depend on the odorants. Note that there are experimental …ndings showing that the response of the same OSN to di¤erent odorants may exhibit di¤erent Hill coe¢ cients (see [7]). Then, starting from Eq. (2) 1 for the odorants U and V

F _U (U ) = F max

1 + 1

n

1 + K U

U

n

; F _V (V ) = F max

1 + 1

n

1 + K V

V

n

; (12)

we propose the following model for binary mixture F _{U +V} (U; V ) = F max

1 + 2 6 6 4

1 + U K U

+ V K V

U

U K U

+

V K V

3 7 7 5

n

(13)

where

n U +V = n U U K V U + n V V K U V

U K V U + V K U V : (14)

Equation (13) is a function of two compounds in concentration U and V: How- ever, if U = rV where r is a constant, we obtain

F _max 1 + 1

n 1 + K V

n (15)

where the parameters = (r;

; _V ; K _U ; K _V ) ; K = K (r; K _U ; K _V ) and n = n (r; n _U ; n _V ;

; _V ; K _U ; K _V ) assume the form

= ^U rK V + V K U

K U + rK V

; K = K U K V

K U + rK V

; n = rn U U K V + n V V K U

r U K V + V K U

: (16) We note that if n U = n V ; then (13) and (15) become (2) 2 and (7), respec- tively.

The asymptotic value of function (15) is F _max 1 + 1

n

=

n F _max

1 + ⁿ ; (17)

and the concentration of V at half maximun response

n F max

2 (1 + ⁿ ) is K

p

2 + ⁿ 1 (18)

Model for mixture of N odorants

Equation (13) can be easily generalized to a model for mixture of N odor- ants

F M IX (U 1 ; :::; U N ) = F max

1 + 2 6 6 6 6 4

1 + X N i=1

U _i K U

X N i=1

Ui

U i

K _U

3 7 7 7 7 5

n

(19)

where

n M IX = X N

i=1

n U

U i

K _U

X N

i=1

Ui

U _i K U

: (20)

Similarly, setting for all i 2 f1; :::; N 1g ; U i = r _i U _N where r _i are positive constants, we obtain

F M IX (r 1 U N ; :::; U N ) = F max

1 + 1

n

M IX

1 + K M IX

U N

n

(21)

where

M IX =

N X 1 i=1

r i

K _U

⁺ 1 K _U

N X 1 i=1

r i

K _U

⁺ 1 K _U

; K M IX = 1=

N X 1 i=1

r i

K U

+ 1

K U

!

;

n M IX =

N X 1 i=1

n U

r _i K U

+ n U

1 K U

N X 1 i=1

Ui

r _i

K _U

+

1 K _U

(22)

Appendix A.3: Analysis of the proposed mixture model

For applicative purposes, one might be interested in knowing a priori for which mixtures our model (13), (14) predicts inhibition, synergy or suppression, as described at the beginning of Appendix A1. Here we give details for the whole range of concentrations; namely, we look for the three ranges of values of U; K U ; U ; n U ; V; K V ; V ; n V , for which those qualitative behaviours occur.

This problem obviously consists in solving (rather cumbersome) inequalities that arise by comparing the values F U +V (U; V ), F U (C), F V (C) given by (13) and (12) with C := U + V .

To use

r := U=V; C = U + V

instead of the pair U; V is perhaps more meaningful and mathematically simpler.

Because of the inverse formulas U = rC

1 + r ; V = C 1 + r we have no loss, nor substantial bias of information.

Note also that if one gets n _U ; n _V multiplied by the same (positive) constant, the behaviour stays the same. Hence the ranges can be described using the parameter

:= n _U =n _V

instead of the pair (n _U ; n _V ). When n _U = n _V we have the Cruz and Lowe model, which has already been discussed. Hence we assume n _U 6= n V . Moreover, to avoid an unnecessary long description, we assume that U denotes the odorant with the least Hill coe¢ cient (exponent n):

n U < n V ;

that is, < 1. Hence, to test parameters against the list of cases we are going to display, one should keep in mind that if n U > n V , then U and V have to be exchanged (either in the data or in the list).

It will also be convenient to set b _U := K _U + C

U C ; b _V := K _V + C

V C

and to consider the function y = x + x ln x. This function has an absolute minimum (x; y) = (exp( 2); exp( 2)), and it is increasing for x exp( 2).

We shall denote by '(y) the (partial) inverse de…ned for y exp( 2) (with

values x exp( 2)). To better understand the cases in the list below, one

should take into account that y > '(y) > 1 when y > 1 and y < '(y) < 1 when

y < 1.

For the sake of brevity, we shall disregard some edge cases: in the list below we shall always use strict inequalities and we also assume

V K _U K _V (K _U + C) 6= U K _U K _V (K _V + C)

(that is, F U +V (rC=(1 + r); C=(1 + r)) is not constant with respect to r). We also renounce to discuss some of the cases that are more cumbersome to be analysed.

When 1 < b U < b V we have suppression for all ; r (under the standing assumptions 0 < < 1, r > 0).

When b U > 1 and (1 <)' (b U ) < b V < b U : – when

(1 >) > ^V (K U + C) U V C (b U + b U ln b U )

U (K _V + C) _{U V} C (b _U + b _U ln b _U ) (> 0) we have suppression for all r(> 0);

– when

(0 <) ln b _V ln b U

< < ^V (K _U + C) _{U V} C (b _U + b _U ln b _U )

U (K V + C) U V C (b U + b U ln b U ) (< 1) there exists exactly one value r _U > 0 for which

F U +V (r U C=(1 + r U ); C=(1 + r U )) = F U (C) (to be determined numerically) and we have

suppression for (0 <)r < r _U , inhibition for r > r U ; – when

(0 <) ^V (K _U + C) _{U V} C (b _V + b _V ln b _V )

U (K V + C) U V C (b V + b V ln b V ) < < ln b _V ln b U

(< 1) there exists exactly one value r _V > 0 for which

F U +V (r V C=(1 + r V ); C=(1 + r V )) = F V (C) (to be determined numerically) and we have

inhibition for (0 <)r < r _V , suppression for r > r V ; – when

(0 <) < ^V (K _U + C) _{U V} C (b _V + b _V ln b _V )

U (K V + C) U V C (b V + b V ln b V )

we have suppression for all r(> 0).

When b U > 1 and 1 < b V < ' (b U ):

– when

(1 >) > ^V (K U + C) U V C (b U + b U ln b U )

U (K _V + C) _{U V} C (b _U + b _U ln b _U ) (> 0) we have suppression for all r(> 0);

– when

(0 <) ln b V

ln b _U < < ^V (K U + C) U V C (b U + b U ln b U )

U (K _V + C) _{U V} C (b _U + b _U ln b _U ) (< 1) there exists exactly one value r U > 0 for which

F _{U +V} (r _U C=(1 + r _U ); C=(1 + r _U )) = F _U (C) (to be determined numerically) and we have

suppression for (0 <)r < r _U , inhibition for r > r U ; – when

(0 <) < ln b _V ln b U

(< 1) there exists exactly one value r V > 0 for which

F U +V (r V C=(1 + r V ); C=(1 + r V )) = F V (C) (to be determined numerically) and we have

inhibition for (0 <)r < r V , suppression for r > r _V . When b _U > 1 and (0 <)b _V < 1:

– when

(1 >) > ^V (K U + C) U V C (b U + b U ln b U )

U (K _V + C) _{U V} C (b _U + b _U ln b _U ) (> 0) we have suppression for all r(> 0);

– when

(0 <) < ^V (K U + C) U V C (b U + b U ln b U )

U (K _V + C) _{U V} C (b _U + b _U ln b _U ) (< 1) there exists exactly one value r _U > 0 for which

F U +V (r U C=(1 + r U ); C=(1 + r U )) = F U (C)

(to be determined numerically) and we have

suppression for (0 <)r < r U , inhibition for r > r _U . When (0 <)b U < 1 and b V > 1:

– when

(1 >) > ^V (K U + C) U V C (b U + b U ln b U )

U (K _V + C) _{U V} C (b _U + b _U ln b _U ) (> 0) we have suppression for all r(> 0);

– when

(0 <) < ^V (K U + C) U V C (b U + b U ln b U )

U (K _V + C) _{U V} C (b _U + b _U ln b _U ) (< 1) there exists exactly one value r U > 0 for which

F _{U +V} (r _U C=(1 + r _U ); C=(1 + r _U )) = F _U (C) (to be determined numerically) and we have

suppression for (0 <)r < r _U , synergy for r > r U .

When b _U < 1 and '(b _U ) < b _V < 1:

– when

(1 >) > ^V (K _U + C) _{U V} C (b _U + b _U ln b _U )

U (K V + C) U V C (b U + b U ln b U ) (> 0) we have suppression for all r(> 0);

– when

(0 <) ln b _V ln b U

< < ^V (K _U + C) _{U V} C (b _U + b _U ln b _U )

U (K V + C) U V C (b U + b U ln b U ) (< 1) there exists exactly one value r _U > 0 for which

F U +V (r U C=(1 + r U ); C=(1 + r U )) = F U (C) (to be determined numerically) and we have

suppression for 0 < r < r U , synergy for r > r _U ;

– when (0 <) < ln b _V = ln b _U (< 1) there exists exactly one value r _V > 0 for which

F U +V (r V C=(1 + r V ); C=(1 + r V )) = F V (C)

(to be determined numerically) and we have

synergy for 0 < r < r V , suppression for r > r V .

When exp( 2) < b _U < 1 and b _U < b _V < ' (b _U ):

– when

(1 >) > ^V (K U + C) U V C (b U + b U ln b U )

U (K _V + C) _{U V} C (b _U + b _U ln b _U ) (> 0) we have suppression for all r(> 0);

– when

(0 <) ln b _V ln b U

< < ^V (K _U + C) _{U V} C (b _U + b _U ln b _U )

U (K V + C) U V C (b U + b U ln b U ) (< 1) there exists exactly one value r _U > 0 for which

F U +V (r U C=(1 + r U ); C=(1 + r U )) = F U (C) (to be determined numerically) and we have

suppression for (0 <)r < r _U , synergy for r > r _U ;

– when

(0 <) ^V (K _U + C) _{U V} C (b _V + b _V ln b _V )

U (K V + C) U V C (b V + b V ln b V ) < < ln b _V ln b U

(< 1)

there exists exactly one value r _V > 0 for which

F U +V (r V C=(1 + r V ); C=(1 + r V )) = F V (C) (to be determined numerically) and we have

synergy for (0 <)r < r _V , suppression for r > r _V ; – when

(0 <) < ^V (K _U + C) _{U V} C (b _V + b _V ln b _V )

U (K V + C) U V C (b V + b V ln b V ) (< 1) we have suppression for all r(> 0).

When (0 <)b U < exp( 2) and exp( 2) < b V < ' (b U ) the analysis is more involved and we omit the result.

When (0 <)b U < b V < exp( 2):

– when

(1 >) > ^V (K U + C) U V C (b V + b V ln b V )

U (K _V + C) _{U V} C (b _V + b _V ln b _V ) (> 0) we have suppression for all r(> 0);

– when (0 <) ln b _V

ln b U

< < ^V (K _U + C) _{U V} C (b _V + b _V ln b _V )

U (K V + C) U V C (b V + b V ln b V ) (< 1) there exists exactly one value r V > 0 for which

F U +V (r V C=(1 + r V ); C=(1 + r V )) = F V (C) (to be determined numerically) and we have

inhibition for (0 <)r < r V , suppression for r > r V ; – when

(0 <) ^V (K U + C) U V C (b U + b U ln b U )

U (K V + C) U V C (b U + b U ln b U ) < < ln b V

ln b U

(< 1) there exists exactly one value r _U > 0 for which

F U +V (r U C=(1 + r U ); C=(1 + r U )) = F U (C) (to be determined numerically) and we have

suppression for 0 < r < r U , inhibition for r > r _U ; – when

(0 <) < ^V (K U + C) U V C (b U + b U ln b U )

U (K _V + C) _{U V} C (b _U + b _U ln b _U ) (< 1) we have suppression for all r(> 0).

When (0 <)b _V < b _U < 1 we have suppression for all (0 <) (< 1), r(> 0).

An important consequence of the above analysis is that the model cannot exhibit synergy for concentrations lower than the concentration that gives the half value response for U (i.e. the odorant with lower Hill coe¢ cient). Indeed, note that none of the cases for which b U > 1 gives synergy. Hence, if for a concentration C we have synergy, it must be b U 1, therefore b U n

1, which leads to

F _U (C) = F _max 1 + b U n

F _max 2 :

Since F max clearly exceeds the maximum response for U , and F U is an increasing

function, we deduce that C must exceed the concentration that gives the half

value response for U .

Appendix A.4: Equations for speci…c experimen- tal protocols

Experiments in Rospars et al., 2008

In [2], for each OSN the odorants U; V; and U + V were tested at the concen- trations

U = M ₀ ^U

2d ; V = M ₀ ^V

2d ; M = U + V = M ₀ ^V 2d + M ₀ ^U

2d ;

where d 2 10 ⁱ⁼⁴ ; i = 0; :::; 11 is the dilution, and M ₀ ^U e M ₀ ^V are the molarities of saturing vapors (see Table 1 in [2]). In this way, in each experiment results

U = rV; M = U + V = (1 + r)V; (23)

where the ratio

r = M ₀ ^U M ₀ ^V remains constant at di¤erent dilutions.

The experimental raw data of dose-response curves for the odorants U; V;

and the binary mixture U + V are given in the form (JP Rospars personal communication)

M ₀ ^U

2d ; F U ; M ₀ ^V

2d ; F V ; M ₀ ^U 2d + M ₀ ^V

2d ; F U +V : (24) By …tting the experimental data (24) for the single odorants U and V we de- termine the dose-response functiones F U and F V (see Eq.(12)), whereas the response F U +V is determined as a function of the mixture concentration M = U + V by mean of Eq.(13) in which V = M=(1 + r), i.e.

F _{U +V} (M ) = 1

1 + 1

n 1 + K

M=(1 + r)

n (25)

where the parameters ; K and n assume the form

= ^U rK V + V K U

K _U + rK _V ; K = K U K V

K _U + rK _V ; n = rn U U K V + n V V K U

r _U K _V + _V K _U : (26)

Experiments in Cruz e Lowe, 2013

Di¤erently to [2], in [3] the mixture is tested at a …xed concentration of one of

the components. In this case, after determining by …tting the numerical values

of the parameters n _U ;

; K _U ; n _V ;

; K _V ; using Eq.(13) we obtain the response

F _{U +V} to concentration of mixture U + V for a …xed value of the odorant V: In

detail, by …xing the concentration C for the odorant V we obtain the response F U +V as a function of concentration of odorant U

F _{U +V} (U; C) = 1

1 + 2 6 6 4

1 + U K U

+ C

K V

U

U K U

+

C K V

3 7 7 5

n

(27)

where

n U +V = n U U K V U + n V V K U C

U K V U + V K U C :

For example, in the cases of the mixture EG + MIEG(1:05) (see Fig.4, bottom right panel) the dose-response function is

F _{U +V} (X; 1:05) = 1

1 + 2 6 6 4

1 + X K U

+ 1:05 K V

U

X K U

+

1:05 K V

3 7 7 5

n U U K V X + 1:05n V V K U U K _V X + 1:05 _V K _U

where X is the concentration of odorant EG.

Performance of the proposed mixture model

In order to evaluate the performance of the proposed mixture model, we use the mean square error (MSE) and the mean absolute percentage error (MAPE) ¹ to compare the experimental data with the predicted data. The MAPE classi…ca- tion leves are shown in Supp. Table S1 [6]

MAPE% Error classi…cation

< 10 Highly accurate

10 20 Good

20 50 Reasonable

> 50 Inaccurate

Supp. Table S1: C lassi…cation levels of M A PE

We recall that

M AP E = 1 n

X

E

P

P

100%

where E

and P

are respectively the experimental and predicted values.

MSE and MAPE for all set of experimental data in Figs.4-5

Fig. 4 lim men lim + men

MSE 5:825 10 ⁴ 2:118 10 ³ 4:583 10 ³

MAPE 10:273 9:972 18:484

cam men cam + men

MSE 1:953 10 ⁴ 2:118 10 ³ 1:275 10 ³

MAPE 3:632 9:972 11:687

cam lim cam + lim

MSE 7:005 10 ⁴ 9:548 10 ⁴ 3:230 10 ²

MAPE 9:115 4:857 27:570

lim lyr lim + lyr

MSE 7:854 10 ⁴ 1:059 10 ³ 1:155 10 ²

MAPE 10:545 29:615 39:598

eg mieg eg + mieg(1:05)

MSE 1:638 10 ³ 3:089 10 ⁴ 5:340 10 ³

MAPE 18:398 4:079 8:431

eg mieg eg + mieg(6:45)

MSE 1:638 10 ³ 3:089 10 ⁴ 1:382 10 ³

MAPE 18:398 4:079 4:228

Suppl. Table S2: M SE and M A PE for the exp erim ents in Fig.4

Fig. 5 lim men lim + men lim + men corrected ( 0:06 log ₁₀ M)

MSE 1:728 10 ³ 4:808 10 ⁵ 3:840 10 ² 3:099 10 ²

MAPE 7:407 2:096 55:957 31:644

eva lyr eva + lyr eva + lyr corrected (0:026 log ₁₀ M)

MSE 7:403 10 ³ 7:603 10 ⁴ 2:153 10 ² 3:645 10 ²

MAPE 10:938 13:810 33:904 19:270

cit lil cit + lil cit + lil corrected ( 0:1 log ₁₀ M)

MSE 3:007 10 ³ 2:230 10 ³ 2:449 10 ² 2:273 10 ²

MAPE 10:174 12:978 27:043 15:988

Suppl. Table S3: M SE and M A PE for the exp erim ents in Fig.5

Appendix A.5

Our purpose in this section is to study from a mathematical point of view the OSN responses modelled by (12) and (13).

Consider …rst the following function

F _{(n; ;s)} (X) = F max

1 + 1 + sX sX

n ; (28)

where the variable X ranges in the real positive numbers, and the parameters n; ; s are positive as well.

Note that when s = 1=K and X is an odorant concentration (U; V; :::) we get (12). In this way OSN responses are labeled by the triples (n; ; s);

(n; ; s) ! F (n; ;s) (X) (29)

and the odor response space is

ORS = 8 >

> <

> >

:

F _max 1 + 1 + sX

sX

n

9 >

> =

> >

;

n; ;s 2R

: (30)

Let us introduce two composition laws (n; ; s) (n ⁰ ; ⁰ ; s ⁰ ) = n s + n ^{0 0} s ⁰

s + ⁰ s ⁰ ; s + ⁰ s ⁰

s + s ⁰ ; s + s ⁰ ;

> (n; ; s) = (n; ; s); 2 R ⁺ :

(31)

We remark that

(n; ; s) (n ⁰ ; ⁰ ; s ⁰ ) identi…es a mixture of two odorants (n; ; s) and (n ⁰ ; ⁰ ; s ⁰ );

i.e.

(n; ; s) (n ⁰ ; ⁰ ; s ⁰ ) ! F (n; ;s) (n

;

;s

) = F max

1 + 1 + (s + s ⁰ ) X ( s + ⁰ s ⁰ ) X

n s + n ^{0 0} s ⁰ s + ⁰ s ⁰

> (n; ; s); 2 R + identi…es the odorant (n; ; s) at concentration X;

i.e.

> (n; ; s) ! F max

1 + ^{1+s( X)} _{s( X)} ⁿ

:

The practical meaning of the above operations is rather simple: to mix odorants (n; ; s), (n ⁰ ; ⁰ ; s ⁰ ) at a …xed ratio r (U = rV ) is equivalent to consider

r

r + 1 > (n; ; s) 1

r + 1 > (n ⁰ ; ⁰ ; s ⁰ ) ;

that is, a weighted mean in term of the operations. Some natural properties can easily be checked: for instance, is commutative and associative. This means, obviously, that a mixture of several odorants does not depend on the way one mixes them, but only on the reciprocal ratios. On the other hand, we can not have ‘opposite odorants’, because of the positivity constraints. The latter is basically the only obstacle that prevents our operations from de…ning a vector space structure. To understand why, let us recall that to de…ne a …nite- dimesional vector space structure over the real numbers is the same as to de…ne a one-to-one map onto R ⁿ (actually, an equivalence class of them): the condition to be ful…lled is that the structural operations must correspond, via the map, with the usual addition and scaling of vectors. We shall see below that such a map can be constructed, with the constraint of not being onto R ⁿ (n = 3).

The space (ORS; ; >) embeds into the standard topological vector space R ³ by means of the map

(n; ; s) 7! (n s; s; s): (32)

Indeed from (32) we obtain that

(n; ; s) (n ⁰ ; ⁰ ; s ⁰ ) 7! (n s; s; s) + (n ^{0 0} s ⁰ ; ⁰ s ⁰ ; s ⁰ ) (33)

> (n; ; s) 7! (n s; s; s) (34) The …rst simple qualitative conclusion suggested from the mathematical point of view is that no privileged odorants arise in the odor space. Indeed, the only ‘special’ element in R ³ is the zero triple, which is out of the range of the embedding (i.e. K lies at in…nity). The special status of the zero triple can be justi…ed by saying that any automorphism (i.e. transformations preserving the mathematical structure) must transform the zero triple into itself. On the contrary, any two nonzero triples can be transformed into each other by some automorphism, and are henceforth equivalent from a structural (mathematical) point of view.

Another interesting feature carried by the vector space structure of R ³ is that the choice of three independent vectors (i.e., a basis) su¢ ces to generate the whole space. The map (32) could be used to carry a similar feature on the odor space,

( ₁ > (n 1 ; ₁ ; s ₁ )) ( ₂ > (n 2 ; ₂ ; s ₂ )) ( ₃ > (n 3 ; ₃ ; s ₃ )) 7!

X 3 i=1

i (n _{i i} s _i ; _i s _i ; s _i ) ;

(35)

but with some cautions. Indeed, the odor space is mapped only on the positive octant R ³ + , because only positive scalars are allowed in order to have a physically meaningful combinations

( ₁ > (n 1 ; ₁ ; s ₁ )) ( ₂ > (n 2 ; ₂ ; s ₂ )) ( ₃ > (n 3 ; ₃ ; s ₃ )) : (36) Thus, let us choose three vectors

e ₁ = (n _{1 1} s ₁ ; ₁ s ₁ ; s ₁ ) ; (37) e 2 = (n 2 2 s 2 ; 2 s 2 ; s 2 ) ;

e ₃ = (n _{3 3} s ₃ ; ₃ s ₃ ; s ₃ ) ;

in the positive octant R ³ + , which constitute a basis provided that

1 2 (n 1 n 2 ) 1 3 (n 1 n 3 ) + 2 3 (n 2 n 3 ) 6= 0: (38) Every vector (n s; s; s) 2 R ³ + is uniquely obtained as a linear combination of (37), i.e.

8(n s; s; s) 9! ( ¹ ; 2 ; 3 ) : (n s; s; s) = X 3 i=1

i (n i i s i ; i s i ; s i ) (39) The scalars in (39) must be positive and can explicitly be determined as follows:

1 = s [ n( 2 3 ) + 2 n 2 ( 3 ) + 3 n 3 ( 2 )]

s ₁ > 0; (40)

2 = s [ n ( 3 1 ) + 1 n 1 ( 3 ) + 3 n 3 ( 1 )]

s ₂ > 0;

3 = s [ n( 1 2 ) + 1 n 1 ( 2 ) + 2 n 2 ( 1 )]

s 3

> 0:

Hence, in ORS, an odorant (n; ; s) is a combination of F _(n

_;

_;s

₎ = F _max

1 + 1 + s ₁ X

1 s ₁ X

n

;

F _(n

_;

_;s

₎ = F max

1 + 1 + s ₂ X

2 s 2 X

n

;

F _(n

_;

_;s

₎ = F max

1 + 1 + s 3 X

3 s ₃ X

n

by means of positive scalars 1 ; 2 ; 3 ;if and only if (37) e (40) are ful…lled. Due to the restriction (40), no triple can cover the whole positive octant.

In order to visualize the generation procedure in the odor response space

ORS, we …rst point out that, from a mathematical viewpoint, two di¤erent

odorants with the same parameters (n; ) behave as if they were the same at di¤erent concentrations. Hence ORS can be e¤ectively represented in the plane (n; ) as its …rst quarter. Combinations in ORS corresponds to convex combi- nations through the map

(n; ) 7! (n ; ): (41)

For instance, let us consider the basis

(n ₁ ; ₁ ; s ₁ ) : n ₁ = 0:1; ₁ = 0:1; s ₁ > 0 (42) (n 2 ; 2 ; s 2 ) : n 2 = 0:1; 2 = 18; s 2 > 0

(n 3 ; 3 ; s 3 ) : n 3 = 18 3 = 0:1; s 3 > 0:

In Fig. 1 is represented the region it generates.

Figure 1

It is a triangle (convex hull of three points) in the space (n ; );deformed according to the map (41).

Suppose now that odorant parameters (n; ; s) are constrained as follows n 2 [n ^min ; n max ] ; 2 [ ^min ; max ] ; s > 0: (43) It is pretty clear that a triple can not generate all of them. However, four suitable odorants will su¢ ce. Indeed, the map (41) tranforms lines of type n =const and =const into lines in the plane (n ; ): Hence the convex hull given by the four vertices

(n _min ; _min ; s ₁ ) ; s ₁ > 0 (n _min ; _max ; s ₂ ) ; s ₂ > 0 (n _max ; _min ; s ₃ ) ; s ₃ > 0 (n max ; max ; s 4 ) ; s 4 > 0

(44)

covers every odorant F _{(n; ;s)} verifying (43).

We remark that if we consider the following two set of three vertices (44)

B ¹ = 8 <

:

(n _min ; _max ; s ₂ ) ; s ₂ > 0 (n _max ; _min ; s ₃ ) ; s ₃ > 0 (n _max ; _max ; s ₄ ) ; s ₄ > 0

9 =

; ; B ² = 8 <

:

(n _min ; _min ; s ₁ ) ; s ₁ > 0 (n _min ; _max ; s ₂ ) ; s ₂ > 0 (n _max ; _min ; s ₃ ) ; s ₃ > 0

9 =

; ; each of them satis…es condition (38). Infact, since n _min < n _max ; and _min <

max , we obtain for B 1 and B 2 ; respectively

1 max min (n _min n _max ) _max ² (n _min n _max ) + _{min max} (n _max n _max )

= max (n min n max ) ( min max ) 6= 0;

2 min max (n min n min ) min min (n min n max ) + max min (n min n max )

= _min (n _min n _max ) ( _min + _max ) 6= 0

In this way, the set B 1 covers every odorant F _{(n; ;s)} verifying the following constraints

min < < max ; ^min n max ( max ) + max n min ( min )

( _{max min} ) < n < n max ; whereas the set B ² covers every odorant F _{(n; ;s)} verifying the constraints

min < < _max ; n _min < n < ^min n max ( max ) + max n min ( min )

( _{max min} ) ;

The curve of equation

min < < max ; n = ^min n max ( max ) + max n min ( min ) ( _{max min} )

can be covered by all the four odorants.

For example, if we suppose that

n 2 [0:1; 18] ; 2 [0:1; 18] ; s > 0; (45) we obtain that the convex hull of vertices

(0:1; 0:1; s 1 ) ; s 1 > 0 (0:1; 18; s 2 ) ; s 2 > 0 (18; 0:1; s 3 ) ; s 3 > 0 (18; 18; s 4 ) ; s 4 > 0

(46)

covers every odorant F _{(n; ;s)} verifying (45). In fact, if we consider the following two triples

(0:1; 18; s 2 ) ; s 2 > 0 (18; 0:1; s 3 ) ; s 3 > 0 (18; 18; s 4 ) ; s 4 > 0

;

(0:1; 0:1; s 1 ) ; s 1 > 0

(0:1; 18; s 2 ) ; s 2 > 0

(18; 0:1; s 3 ) ; s 3 > 0

we easily verify that generate the regions in Figs 2 and 3, respectively.

Figure 2 Figure 3

We remark that the union of the regions in Figs.2-3 is able to cover the entire rectangle [0:1; 18] [0:1; 18] in the phase plane (n; ) except for the curve of equation

= 9

5n ; 0:1 < n < 18: (47)

Therefore, to include the curve (47) we use the four vectors (46). In particular, setting

s = s 1 = s 2 = s 3 = s 4 = 1;

we obtain that n; 9

5n ; 1 = ( 1 > (0:1; 0:1; 1)) ( 2 > (0:1; 18; 1)) ( 3 > (18; 0:1; 1)) ( 4 > (18; 18; 1)) in which

2 = 18 n 179n

1

180 ; 3 = 18(10n 1)

179n ¹ ; 4 = ¹

180 ; and

1 <

8 >

> <

> >

:

180n 18 179n ; if 1

10 < n 181 20 ; 3240 180n

179n ; if 181

20 < n < 18:

:

This shows that a point of the curve (47) (except the endpoints) can be

obtained in in…nitely many ways as a combination of the four vectors with

positive coe¢ cients, and in exactly one way as a combination of the two vectors

(0:1:18; 1), (18; 0:1; 1) with positive coe¢ cients.

Appendix A.6

We consider a set of four OSN responses to concentration of odorants U; V

OSN odorant U odorant V

1 n _U = 11:5; _U = 4:5; K _U = 8 10 ⁵ n _V = 6:7; _V = 10:5; K _V = 10 ⁴ 2 n _U = 13:7; _U = 3:5; K _U = 8 10 ⁵ n _V = 3:4; _V = 11:4; K _V = 10 ⁴ 3 n U = 14:5; U = 2:8; K U = 8 10 ⁵ n V = 4:3; V = 12:6; K V = 10 ⁴ 4 n U = 12:5; U = 1:9; K U = 8 10 ⁵ n V = 5:9; V = 13:9; K V = 10 ⁴ (48) Owing to (15), the OSN responses to binary mixture U + V for r = 1 are

OSN binary mixture U + V 1 n = 8:374; = 7:167; K = 4:4 10 ⁵ 2 n = 6:256; = 7:011; K = 4:4 10 ⁵ 3 n = 6:517; = 7:155; K = 4:4 10 ⁵ 4 n = 6:863; = 7:233; K = 4:4 10 ⁵

(49)

Since the odorant parameters (n; ; K) are constrained as follows n 2 [0:1; 18] ; 2 [0:1; 18] ; s > 0;

it is easy to verify that the OSN responses to odorants U; V and U + V are located into the region

> 9

5n ; 0:1 < n < 18;

of the plane (n; ). Then, this region is generated by the basis (0:1; 18; s 1 ) ; s 1 > 0

(18; 0:1; s 2 ) ; s 2 > 0 (18; 18; s 3 ) ; s 3 > 0

(50)

and setting s 1 = s 2 = s 3 = 10 ⁵ ; we obtain that all the OSN responses to odorants U; V and U + V are combination of vectors (50), i.e.

1 > 0:1; 18; 10 ⁵ 2 > 18; 0:1; 10 ⁵ 3 > 18; 18; 10 ⁵

according to the following values of ₁ ; ₂ ; ₃

OSN scalars odorant U odorant V mixture U + V 1

1 2 3

0:011 0:094 0:019

0:037 0:042 0:021

0:048 0:136 0:041

2

1 2 3

0:006 0:101 0:018

0:052 0:037 0:011

0:057 0:138 0:029

3

1 2 3

0:004 0:106 0:015

0:053 0:030 0:016

0:057 0:136 0:031

4

1 2 3

0:004 0:112 0:008

0:052 0:023 0:025

0:056 0:135 0:033

In the above example we considered for simplicity the mixture U + V (at

the same concentration) for all OSN. In general, di¤erent OSNs may lead to

mixtures of U and V at di¤erent concentrations.

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